Let $f ( x ) = ( \tan x ) ^ { 3 / 2 } - 3 \tan x + \sqrt{\tan x}$. Consider the three integrals $I _ { 1 } = \int_0^1 f ( x ) \, dx$; $I _ { 2 } = \int_{0.3}^{1.3} f ( x ) \, dx$ and $I _ { 3 } = \int_{0.5}^{1.5} f ( x ) \, dx$. Then,
(a) $I _ { 1 } > I _ { 2 } > I _ { 3 }$
(b) $I _ { 2 } > I _ { 1 } > I _ { 3 }$
(c) $I _ { 3 } > I _ { 1 } > I _ { 2 }$
(d) $I _ { 1 } > I _ { 3 } > I _ { 2 }$
(D) $\int f(x)\,dx = 2\sqrt{\tan x} + (1/\sqrt{2})\ln|(\tan x - \sqrt{2\tan x}+1)/(\tan x + \sqrt{2\tan x}+1)| + 3\ln|\cos x| + C$. Evaluating at the given limits gives $I_1 > I_3 > I_2$.
Let $f ( x ) = ( \tan x ) ^ { 3 / 2 } - 3 \tan x + \sqrt{\tan x}$. Consider the three integrals $I _ { 1 } = \int_0^1 f ( x ) \, dx$; $I _ { 2 } = \int_{0.3}^{1.3} f ( x ) \, dx$ and $I _ { 3 } = \int_{0.5}^{1.5} f ( x ) \, dx$. Then,\\
(a) $I _ { 1 } > I _ { 2 } > I _ { 3 }$\\
(b) $I _ { 2 } > I _ { 1 } > I _ { 3 }$\\
(c) $I _ { 3 } > I _ { 1 } > I _ { 2 }$\\
(d) $I _ { 1 } > I _ { 3 } > I _ { 2 }$